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Thread: Integrating factors

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    Integrating factors

    So, some guidance would be helpful.

    $\displaystyle ydx+(2xy-e^{-2y})dy = 0
    $

    $\displaystyle \frac{1-2y}{-y}$

    $\displaystyle e^{^-{\int \frac{1}{y}-2}} $

    $\displaystyle \frac{e^{2y}}{y}$ is the integrating factor.

    And upon solving, the answer is: $\displaystyle C = xe^{2y}-ln|y|$?
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    $\displaystyle y\,dx + (2xy - e^{-2y})\,dy = 0$

    $\displaystyle y\,\frac{dx}{dy} + 2xy - e^{-2y} = 0$

    $\displaystyle \frac{dx}{dy} + 2x - \frac{e^{-2y}}{y} = 0$

    $\displaystyle \frac{dx}{dy} + 2x = \frac{e^{-2y}}{y}$.


    The integrating factor is $\displaystyle e^{\int{2\,dy}} = e^{2y}$

    So multiplying through by the integrating factor gives

    $\displaystyle e^{2y}\,\frac{dx}{dy} + 2e^{2y}x = \frac{1}{y}$

    $\displaystyle \frac{d}{dy}(e^{2y}x) = \frac{1}{y}$

    $\displaystyle e^{2y}x = \int{\frac{1}{y}\,dy}$

    $\displaystyle e^{2y}x = \ln{|y|} + C$

    $\displaystyle x = e^{-2y}\ln{|y|} + Ce^{-2y}$.
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    If I had solved for x at the end, I would've gotten the same answer. Is my integrating factor incorrect?
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    Quote Originally Posted by A Beautiful Mind View Post
    If I had solved for x at the end, I would've gotten the same answer. Is my integrating factor incorrect?
    I don't see how you got your Integrating Factor to be honest...

    But your final answer looks correct.
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    I applied a formula.

    I don't know how to do the subscript things on latex but...

    $\displaystyle \frac{My-Nx}{-M}$
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